ebook img

Inversion of spherical means and the wave equation in even dimensions PDF

0.57 MB·English
Save to my drive
Quick download
Download
Most books are stored in the elastic cloud where traffic is expensive. For this reason, we have a limit on daily download.

Preview Inversion of spherical means and the wave equation in even dimensions

Inversion of spherical means and the wave equation in 7 0 even dimensions 0 2 n a J David Finch Markus Haltmeier 5 1 Department of Mathematics Department of Computer Science ] Oregon State University Universit¨at Innsbruck P A Corvallis, OR 97331 Technikerstraße 21a . h fi[email protected] A-6020 Innsbruck, Austria t a [email protected] m [ Rakesh 1 Department of Mathematical Sciences v 6 University of Delaware 2 4 Newark, DE 19716 1 [email protected] 0 7 0 / h t January 15, 2007 a m : v i X Keywords: spherical means, wave equation, thermoacoustic tomography r a AMS subject classifications: 35R30, 35L05, 35Q05, 92C55, 65R32 Abstract We establish inversion formulas of the so called filtered back-projection type to recover a function supported in the ball in even dimensions from its spherical means over spheres centered on the boundary of the ball. We also find several formulas to recover initial data of the from (f,0) (or (0,g)) for the free space wave equation in even dimensions from the trace of the solution on the boundary of the ball, provided the initial data has support in the ball. 1 1 Introduction and Statement of Results The problem of determining a function from a subset of its spherical means has a rich history in pure and applied mathematics. Our interest in the subject was provoked by the newmedicalimagingtechnologiescalledthermoacousticandphotoacoustictomography. The idea behind these [10, 16] is to illuminate an object by a short burst of radiofrequency or optical energy which causes rapid (though small in magnitude) thermal expansion which generates an acoustic wave. The acoustic wave can be measured on the periphery or in the exterior of the object. The inverse problem we consider is to find the distribution of the absorbed energy throughout the body. This is of interest, since the amount of energy absorbed at different points may be diagnostic of disease or indicative of uptake of probes tagged to metabolic processes or gene expression [9]. For a more thorough discussion of the modelling and biomedical applications, the reader is referred to the recent survey [17]. If the illuminating energy is impulsive in time, the propagation may be modelled as an initial value problem for the wave equation. The problem of recovering the initial data of a solutionofthe wave equation fromthevalueofthe solutiononthe boundaryofadomainisof mathematical interest in every dimension, but for the application to thermo-/photoacoustic tomography it would appear that the three dimensional case is the only one of interest, since sound propagation is not confined to a lower dimensional submanifold. However, there exist methods of measuring the generated wave field which do not rely on point measurements of the sort that would be generated by an (idealized) acoustic transducer. In particular, integrating line detectors, which have been studied in [3, 14], in effect compute the integral of the acoustic wave field along a specified line. In this paper, we work under the assumption that the speed of sound, c, is constant throughout the body, and since the x-ray transform in a given direction of a solution of the three dimensional wave equation is a solution of the two dimensional wave equation, the problem is transformed. If a circular array of line detectors is rotated around an axis orthogonal to the direction of the line detectors [7, 14], then for each fixed rotation angle the measurement provides the trace of the solution of the two dimensional wave equation on the circle corresponding to the array. The initial data of this two dimensional problem is the x-ray transform of the three dimensional initial data. If the inital data can be recovered in the disk bounded by the detector array and assuming that the projection of the object to be imaged lies in this disk, then the problem of recovering the three dimensional initial data is reduced to the inversion of the x-ray transform in each plane orthogonal to the axis of rotation. One such two dimensional problem is illustrated in Figure 1. To our knowledge, the first work to tackle the problem of recovering a function from its circular means with centers on a circle was [13], whose author was interested in ultrasound reflectivity tomography. He found an inversion method based on harmonic decomposition and for each harmonic, the inversion of a Hankel transform. This method has been the basis for most subsequent work on exact inversion of circular means. The inversion of the Hankel 2 line detector e1 p S ∈ R2 rotation Figure1: Principle of thermoacoustic tomography with integrating line detectors. A cylindrical array of line detectors records the acoustic field and is rotated around the axis e . For fixed rotation angle the array outputs the x-ray transform (projection along straight 1 lines) of the solution of the wave equation restricted to lines passing through the boundary S of the disk. The initial condition is given by the x-ray transform of the initially induced pressure restricted to lines orthogonal to the base of the cylinder. transform involves a quotient of a Hankel transform of a harmonic component of the data and a Bessel function. That this quotient be well-defined turns out to be a condition on the range of the circular mean transform [2]. See also [1] for range results on the spherical mean transform on functions supported in a ball in all dimensions, and [6] for range results for the wave trace map for functions supported in the ball in odd dimensions. In the work of the first and third authors with Sarah Patch [5], several formulas were found to recover a smooth function f with support in the closure B of the open ball B Rn ⊆ fromthetraceofthesolutionofthewave equationontheproduct ∂B [0,diam(B)]provided × 3 that the space dimension is odd. Specifically, if u is the solution of the initial value problem u ∆u = 0, in Rn [0, ) (1) tt − × ∞ u(.,t = 0) = f(.), u (.,t = 0) = 0, (2) t where f is smooth and has support in the B, then several formulas were found to recover f from u(p,t) for p S := ∂B and t R+. ∈ ∈ The first and third authors tried, at that time, to extend the method to even dimensions, but did not see a way. Recently, the second author tried numerical experiments using a two dimensional analog of one of the inversion formulas and found that it gave excellent reconstructions. This prompted our re-examination of the problem. Among the results of this paper is a proof of the validity of this formula. To describe our results, we introduce some notation. The spherical mean transform M is defined by 1 ( f)(x,r) = f(x+rθ)dS(θ) (3) M Sn 1 | − | ZSn−1 for f C (Rn) and (x,r) Rn [0, ). In this expression, Sn 1 denotes the area of the ∞ − ∈ ∈ × ∞ | | unit sphere Sn 1 in Rn and dS(θ) denotes area measure on the sphere. In general, we write − the area measure on a sphere of any radius as dS, except when n = 2 when we write ds. We will denote the (partial) derivative of a function q with respect to a variable r by ∂ q, except r in a few formulas where the subscript notation q is used. At several points we use D to r r denote the operator (∂ u)(r) r (D u)(r) := r 2r acting on smooth (even) functions u with compact support. Moreover, r will be used to denote the operator that multiplies a function u(r) by r. Our first set of results is a pair of inversion formulas for the spherical mean transform in even dimensions. We state and prove these first in dimension two; that is, for the circular mean transform. Theorem 1. Let D R2 be the disk of radius R centered at the origin, let S := ∂D denote 0 ⊂ the boundary circle, and let f C (R2) with suppf D. Then, for x D, ∞ ∈ ⊂ ∈ 1 2R0 f(x) = ∆ r( f)(p,r)log r2 x p 2 drds(p) (4) x 2πR M −| − | 0 ZSZ0 (cid:12) (cid:12) and (cid:12) (cid:12) 1 2R0 f(x) = (∂ r∂ f)(p,r)log r2 x p 2 drds(p). (5) r r 2πR M −| − | 0 ZSZ0 (cid:12) (cid:12) (cid:12) (cid:12) 4 In Theorem 1, ∂ r∂ f denotes the composition of ∂ , r, ∂ and applied to f. r r r r M M The same convention will be used throughout the article to denote the composition of any operators. While f has a natural extension to the negative reals as an even function, we instead M take the odd extension in the second variable. Then formula (5) has the following corollary: Corollary 1. With the same hypotheses as in Theorem 1, and f extended as an odd M function in the second variable r, f can be recovered for x D by ∈ 1 2R0 (r∂ f)(p,r) r f(x) = M drds(p), (6) 2πR x p r 0 ZSZ−2R0 | − |− and 1 2R0 (∂ f)(p,r) r f(x) = x p M drds(p), (7) 2πR | − | x p r 0 ZS Z−2R0 | − |− where the inner integrals are taken in the principal value sense. These forms are very close to the standard inversion formula for the Radon transform in the plane [12, Eq. (2.5)]. In higher even dimensions we prove a similar pair of results. Theorem 2. Let B Rn, n > 2 even, be the ball of radius R centered at the origin, let 0 ⊂ S := ∂B be the boundary of the ball, set c = ( 1)(n 2)/22((n 2)/2)!πn/2 = ( 1)(n 2)/2[((n 2)/2)!]2 Sn 1 , n − − − − − − − | | and let f C (Rn) have support in B. Then, for x B, ∞ ∈ ∈ 1 2R0 f(x) = ∆ log r2 x p 2 (rDn 2rn 2 f)(p,r)drdS(p), (8) c R x −| − | r− − M n 0 ZSZ0 2 2R0 (cid:12) (cid:12) f(x) = log r2 (cid:12) x p 2 (rD(cid:12) n 1rn 1∂ f)(p,r)drdS(p). (9) c R −| − | r− − rM n 0 ZSZ0 (cid:12) (cid:12) (cid:12) (cid:12) Recently, Kunyansky [11] has also established inversion formulas of the filtered back- projection type for the spherical mean transform. His method and results appear to be very different than ours. For some results, it will be more convenient to use the wave equation (1) with initial condition u(.,t = 0) = 0, u (.,t = 0) = f(.), (10) t 5 It is obvious that the solution of (1) with initial values (2) is the time derivative of the solution of (1) with initial values (10). We denote by the operator which takes smooth P initial data with support in B to the solution of (1), (10) restricted to S [0, ) and by × ∞ W the operator taking f to the solution of (1), (2) restricted to S [0, ). These operators are × ∞ simply related by = ∂ . An explicit representation for comes from the well-known t W P P formula [4] 1 t u(p,t) = ∂n 2 r(t2 r2)(n 3)/2( f)(p,r)dr. (11) (n 2)! t− − − M − Z0 giving the solution of the initial value problem (1), (10), in dimension n 2. We denote by ≥ and = ∂ the formal L2 adjoints of and mapping from smooth functions ∗ ∗ ∗ t P W −P P W u C (S [0, )) with sufficient decay in the second variable. An explicit expression for ∞ ∈ × ∞ u will be given in Section 3. ∗ P We have two types of inversion results for the wave equation. The first type is based on the inversion results for the spherical mean transform, since the spherical mean transform itself can be recovered from the solution of the wave equation by solving an Abel type equation. In dimension two, this approach yields the following result. Theorem 3. Let D R2 be the open disc with radius R and let S := ∂D denote the 0 ⊂ boundary circle. Then there exists a kernel function K : [0,2R ]2 R such that for any 0 → f C (R2) with support in D and any x D ∞ ∈ ∈ 1 2R0 f(x) = ∆ ( f)(p,t)K(t, x p )dtds(p). (12) R π2 x W | − | 0 ZSZ0 An analytic expression for K will be given in Section 3. Theorem 3 provides inversion formulas of the filtered back-projection type for recon- struction of f from ( f)(p,t) = (∂ f)(p,t) using only data with t [0,2R ], despite the t 0 W P ∈ unbounded support of f and f in t. W P The second type of inversion results holds in all even dimensions and takes the following form. Theorem 4. Let f be smooth and supported in closure of the ball B of radius R in R2m, 0 and let f and f be as above. Then for x B P W ∈ 2 f(x) = t∂2 f (x), (13) −R P∗ t P 0 2 (cid:0) (cid:1) 2 f(x) = ( t f)(x) = ( ∂ t∂ f)(x). (14) ∗ ∗ t t R W W −R P P 0 0 6 We will prove (13) in dimension n = 2m = 2 directly. The higher dimensional case of (13), and (14) in all dimensions, are consequences of the following trace identities, relating the L2 inner product of the initial data to the weighted L2 inner product of the traces of the solutions of the wave equation. Theorem 5. Let f,g be smooth and supported in the ball B of radius R , in R2m with 0 m 1, let S := ∂B, and let u (resp. v) be the solution of the initial value problem (1), (10) ≥ with initial value f (resp. g). Then 2 ∞ f(x)g(x)dx = tu (p,t)v(p,t)dtdS(p), (15) tt −R ZB 0 ZS Z0 2 ∞ f(x)g(x)dx = tu (p,t)v (p,t)dtdS(p). (16) t t R ZB 0 ZS Z0 In the proof of this theorem, (15) for n = 2 follows from (13) for n = 2, while (15) in higher even dimensions is derived from the n = 2 case; (16) is a consequence of (15) in all dimensions. We remark that these identities were already proved in [5] for odd dimensions, and so they hold for all dimensions. Section 2 is devoted to the proof of the inversion formulas for the spherical mean trans- form, that is, Theorems 1, 2, and Corollary 1. Section 3 treats the wave equation and contains the proofs of Theorems 3, 4, and 5. This is followed by a section reporting on the implementation of the various reconstruction formulas of the preceding sections and results of numerical tests, in dimension two. 2 Spherical Means In this section we prove the Theorems related to the inversion from spherical means and Corollary 1. We begin by establishing an elementary integral identity, which is the key to the results in this paper. Proposition 2.1. Let D R2 be the disk of radius R , and let S = ∂D be the boundary 0 ⊆ circle. Then for x, y D with x = y, ∈ 6 log x p 2 y p 2 ds(p) = 2πR log x y +2πR logR . (17) 0 0 0 | − | −| − | | − | ZS (cid:12) (cid:12) (cid:12) (cid:12) Proof. Let x = y both lie in D and let I denote the integral on the left on (17). Expanding 6 7 the argument of the logarithm as x+y p x y x p 2 y p 2 = 2R x y − , 0 | − | −| − | | − | 2R − R · x y (cid:12)(cid:18) 0 0(cid:19) | − |(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) setting e := xx−yy(cid:12), and writing p =(cid:12)R0θ for θ ∈ S1,(cid:12)(cid:12)we have (cid:12)(cid:12) | − | I = 2πR log(2R x y )+R log e θ a dθ, (18) 0 0 0 | − | | · − | ZS1 where x+y x 2 y 2 a = e = | | −| | . 2R · 2R x y 0 0 | − | We note that a < 1. | | Using the parameterization θ = cos(φ)e+sin(φ)e , the integral term on the right of (18) ⊥ has the form 2π R log cosφ a dφ. 0 | − | Z0 Writing a = cosα and using the sum to product trigonometric identity cosφ cosα = − 2sin((φ+α)/2)sin((φ α)/2), this is equal to − − 2π R (log2+log sin((φ+α)/2) +log sin((φ α)/2) )dφ. 0 | | | − | Z0 By periodicity, and two linear changes of variable, this reduces to 2π π R (log2+2log sin(φ/2) )dφ = 2R πlog2+4R logsinudu, 0 0 0 | | Z0 Z0 which is independent of α, and hence of x and y. The latter integral in can be found in tables, and is equal to R πlog2, so the sum is 2πR log2. Substituting in (18) gives the 0 0 − − desired result. Proposition 2.1 is already enough to establish Theorem 1. Proof of Theorem 1. Let f C (R2) be supported in D and let p be any point in ∞ ∈ S = ∂B. Using the definition of f and Fubini’s theorem, we have that M 2R0 1 (r f)(p,r)q(r)dr = f(p+z)q( z )dz, (19) M 2π | | Z0 ZR2 8 foranymeasurablefunctionq providedthattheproductoffunctionsontherightisabsolutely integrable. Applying this with q(r) = log r2 x p 2 and making the change of variables | −| − | | y = p+z gives 2R0 (r )(f)(p,r)log r2 x p 2 drds(p) M −| − | ZSZ0 (cid:12) (cid:12) 1 (cid:12) =(cid:12) f(y)log y p 2 x p 2 dyds(p). 2π | − | −| − | ZS ZR2 (cid:12) (cid:12) Fubini’s theorem again justifies the change of order of integra(cid:12)tion in the iterated(cid:12) integral on the right hand side, and so 1 2πR f(y) log y p 2 x p 2 ds(p)dy = 0 f(y)(log x y +logR )dy 0 2π | − | −| − | 2π | − | ZR2 ZS ZR2 (cid:12) (cid:12) upon application of(cid:12)(17). Recalling t(cid:12)hat for any constant c, 1/(2π)log x y + c is a | − | fundamental solution of the Laplacian in R2, we have 1 2R0 f(x) = ∆ (r f)(p,r)log r2 x p 2 drds(p), x 2πR M | −| − | | 0 ZS Z0 which proves (4). The second formula, (5), has a similar proof. In this case, we use that the spherical means satisfy the Euler-Poisson-Darboux equation [4] 1 (∂2 f)(x,r)+ (∂ f)(x,r) = (∆ f)(x,r) = ( ∆f)(x,r). r M r rM M M The left hand side of the Darboux equation may be written as (1/r)(∂ r∂ f)(x,r), so the r r M expression on the right of (5) may be rewritten as 1 2R0 (r ∆f)(p,r)log r2 x p 2 drds(p). (20) 2πR M −| − | 0 ZS Z0 (cid:12) (cid:12) Again applying (19), now with the function q(r) =(cid:12)rlog r2 x(cid:12) p 2 and ∆f instead of f, | −| − | | interchanging the order of integration and using (17) shows that the expression (20) is equal to 1 ∆ f(y)(log x y +logR )dy = f(x), y 0 2π | − | R2 Z 2 since no boundary terms arise in view of the support hypothesis on f. Proof of Corollary 1. Let x D, and let ∈ 2R0 U(p,x) := (∂ r∂ f)(p,r)log r2 x p 2 dr r r M −| − | Z0 (cid:12) (cid:12) (cid:12) (cid:12) 9 denote the inner integral in (5). Taking the support of f into account, writing the logarithm as log r2 x p 2 = log r x p +log r + x p , −| − | | −| − || | | − || and integrating (5) by(cid:12)parts leads t(cid:12)o (cid:12) (cid:12) (r∂ f)(p,r) (r∂ f)(p,r) ∞ r ∞ r U(p,x) = P.V. M dr M dr. − r x p − r + x p Z0 −| − | Z0 | − | Here we have used that the distributional derivative of log r is P.V. 1 as well as an ordinary | | r integration by parts. Therefore (5) implies 1 f(x) = U(p,r)ds(p) 2πR 0 ZS 1 2R0 (r∂ f)(p,r) 1 2R0 (r∂ f)(p,r) r r = − M drds(p)+ − M drds(p), 2πR r x p 2πR r+ x p 0 ZSZ0 −| − | 0 ZSZ0 | − | (21) where the inner integral of the first term on the right is taken in the principal value sense. The odd extension of f, f(p, r) := f(p,r), is smooth on R since f vanishes M M − −M M to infinite order at r = 0 by the support hypothesis on f and (r∂ f)(p,r) is an odd r M function in r. Substituting r = r in the second integral in (21) gives − 1 2R0 (r∂ f)(p,r) 1 0 (r∂ f)(p,r) r r f(x) = − M drds(p)+ − M drds(p) 2πR r x p 2πR r x p 0 ZSZ0 −| − | 0 ZSZ−2R0 −| − | and hence 1 2R0 (r∂ f)(p,r) r f(x) = M drds(p). 2πR x p r 0 ZSZ−2R0 | − |− This is (6). To prove (7), it suffices to write r x p = 1+ | − | x p r − x p r | − |− | − |− in (6) and to note that 2R0 (∂ f)(p,r)dr = 0, by the support hypothesis on f. 2 2R0 rM − R 2.1 Proof of Theorem 2 We have found several proofs of Theorem 2, the extension of Theorem 1 to higher even dimensions. The one we present is based on reduction of the higher dimensional problem to the two dimensional case already established. Another, which is not presented in this article, is based on an extension of (17) to higher dimensions. 10

See more

The list of books you might like

Most books are stored in the elastic cloud where traffic is expensive. For this reason, we have a limit on daily download.